BUFFER SOLUTIONS A guide for A level students

Buffer solutions INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A 2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A 2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at. . . www. argonet. co. uk/users/hoptonj/sci. htm Navigation is achieved by. . . either or clicking on the grey arrows at the foot of each page using the left and right arrow keys on the keyboard

Buffer solutions CONTENTS • What is a buffer solution? • Uses of buffer solutions • Acidic buffer solutions • Alkaline buffer solutions • Buffer solutions - ideal concentration • Calculating the p. H of a buffer solution • Salt hydrolysis • Check list

Buffer solutions Before you start it would be helpful to… • know that weak acids and bases are only partly ionised in solution • be able to calculate p. H from hydrogen ion concentration • be able to construct an equation for the dissociation constant of a weak acid

Buffer solutions - Brief introduction Definition “Solutions which resist changes in p. H when small quantities of acid or alkali are added. ” Acidic Buffer (p. H < 7) made from Alkaline Buffer (p. H > 7) made from a Uses a weak acid + its sodium or potassium salt ethanoic acid sodium ethanoate weak base ammonia Standardising p. H meters Buffering biological systems (eg in blood) Maintaining the p. H of shampoos + its chloride ammonium chloride

Buffer solutions - uses “Solutions which resist changes in p. H when small quantities of acid or alkali are added. ” Definition Biological Uses In biological systems (saliva, stomach, and blood) it is essential that the p. H stays ‘constant’ in order for any processes to work properly. e. g. If the p. H of blood varies by 0. 5 it can lead to unconsciousness and coma Most enzymes work best at particular p. H values. Other Uses Many household and cosmetic products need to control their p. H values. Shampoo Buffer solutions counteract the alkalinity of the soap and prevent irritation Baby lotion Buffer solutions maintain a p. H of about 6 to prevent bacteria multiplying Others Washing powder, eye drops, fizzy lemonade

Acidic buffer solutions - action It is essential to have a weak acid for an equilibrium to be present so that ions can be removed and produced. The dissociation is small and there are few ions. relative concs. NB CH 3 COOH(aq) HIGH CH 3 COO¯(aq) LOW + H+(aq) LOW A strong acid can’t be used as it is fully dissociated and cannot remove H +(aq) HCl(aq) ——> Cl¯(aq) + H+(aq) Adding acid Any H+ is removed by reacting with CH 3 COO¯ ions to form CH 3 COOH via the equilibrium. Unfortunately, the concentration of CH 3 COO¯ is small and only a few H+ can be “mopped up”. A much larger concentration of CH 3 COO¯ is required. To build up the concentration of CH 3 COO¯ ions, sodium ethanoate is added.

Acidic buffer solutions - action It is essential to have a weak acid for an equilibrium to be present so that ions can be removed and produced. The dissociation is small and there are few ions. relative concs. NB CH 3 COOH(aq) HIGH CH 3 COO¯(aq) LOW + H+(aq) LOW A strong acid can’t be used as it is fully dissociated and cannot remove H +(aq) HCl(aq) ——> Cl¯(aq) Adding alkali This adds OH¯ ions which react with H+ ions + H+(aq) + OH¯(aq) H 2 O(l) Removal of H+ from the weak acid equilibrium means that, according to Le Chatelier’s Principle, more CH 3 COOH will dissociate to form ions to replace those being removed. CH 3 COOH(aq) CH 3 COO¯(aq) + H+(aq) As the added OH¯ ions remove the H+ from the weak acid system, the equilibrium moves to the right to produce more H+ ions. Obviously, there must be a large concentration of undissociated acid molecules to be available.

Alkaline buffer solutions - action Alkaline buffer Very similar but is based on the equilibrium surrounding a weak base; AMMONIA relative concs. but one needs ; NH 3(aq) + H 2 O(l) HIGH OH¯(aq) + NH 4+(aq) LOW a large conc. of OH¯(aq) to react with any H+(aq) added a large conc of NH 4+(aq) to react with any OH¯(aq) added There is enough NH 3 to act as a source of OH¯ but one needs to increase the concentration of ammonium ions by adding an ammonium salt. Use AMMONIA (a weak base) + AMMONIUM CHLORIDE (one of its salts)

Buffer solutions - ideal concentration The concentration of a buffer solution is also important If the concentration is too low, there won’t be enough CH 3 COOH and CH 3 COO¯ to cope with the ions added. Summary For an acidic buffer solution one needs. . . large [CH 3 COOH(aq)] - for dissociating into H+(aq) when alkali is added large [CH 3 COO¯(aq)] - for removing H+(aq) as it is added This situation can’t exist if only acid is present; a mixture of the acid and salt is used. The weak acid provides the equilibrium and the large CH 3 COOH(aq) concentration. The sodium salt provides the large CH 3 COO¯(aq) concentration. One uses a WEAK ACID + its SODIUM OR POTASSIUM SALT

Calculating the p. H of an acidic buffer solution Calculate the p. H of a buffer whose [HA] is 0. 1 mol dm-3 and [A¯] of 0. 1 mol dm-3.

Calculating the p. H of an acidic buffer solution Calculate the p. H of a buffer whose [HA] is 0. 1 mol dm-3 and [A¯] of 0. 1 mol dm-3. Ka = [H+(aq)] [A¯(aq)] [HA(aq)]

Calculating the p. H of an acidic buffer solution Calculate the p. H of a buffer whose [HA] is 0. 1 mol dm-3 and [A¯] of 0. 1 mol dm-3. Ka = [H+(aq)] [A¯(aq)] [HA(aq)] re-arrange [H+(aq)] = [HA(aq)] x Ka [A¯(aq)]

Calculating the p. H of an acidic buffer solution Calculate the p. H of a buffer whose [HA] is 0. 1 mol dm-3 and [A¯] of 0. 1 mol dm-3. Ka = [H+(aq)] [A¯(aq)] [HA(aq)] re-arrange [H+(aq)] = [HA(aq)] x Ka [A¯(aq)] from information given [A¯] = 0. 1 mol dm-3 [HA] = 0. 1 mol dm-3

Calculating the p. H of an acidic buffer solution Calculate the p. H of a buffer whose [HA] is 0. 1 mol dm-3 and [A¯] of 0. 1 mol dm-3. Ka = [H+(aq)] [A¯(aq)] [HA(aq)] re-arrange [H+(aq)] = [HA(aq)] x Ka [A¯(aq)] from information given [A¯] = 0. 1 mol dm-3 [HA] = 0. 1 mol dm-3 If the Ka of the weak acid HA is 2 x 10 -4 mol dm-3. [H+(aq)] = 0. 1 x 2 x 10 -4 0. 1 = 2 x 10 -4 mol dm-3

Calculating the p. H of an acidic buffer solution Calculate the p. H of a buffer whose [HA] is 0. 1 mol dm-3 and [A¯] of 0. 1 mol dm-3. Ka = [H+(aq)] [A¯(aq)] [HA(aq)] re-arrange [H+(aq)] = [HA(aq)] x Ka [A¯(aq)] from information given [A¯] = 0. 1 mol dm-3 [HA] = 0. 1 mol dm-3 If the Ka of the weak acid HA is 2 x 10 -4 mol dm-3. [H+(aq)] = 0. 1 x 2 x 10 -4 = 2 x 10 -4 mol dm-3 0. 1 p. H = - log 10 [H+(aq)] = 3. 699

Calculating the p. H of an acidic buffer solution Calculate the p. H of the solution formed when 500 cm 3 of 0. 1 mol dm-3 of weak acid HX is mixed with 500 cm 3 of a 0. 2 mol dm-3 solution of its salt Na. X. Ka = 4 x 10 -5 mol dm-3.

Calculating the p. H of an acidic buffer solution Calculate the p. H of the solution formed when 500 cm 3 of 0. 1 mol dm-3 of weak acid HX is mixed with 500 cm 3 of a 0. 2 mol dm-3 solution of its salt Na. X. Ka = 4 x 10 -5 mol dm-3. Ka = [H+(aq)] [X¯(aq)] [HX(aq)]

Calculating the p. H of an acidic buffer solution Calculate the p. H of the solution formed when 500 cm 3 of 0. 1 mol dm-3 of weak acid HX is mixed with 500 cm 3 of a 0. 2 mol dm-3 solution of its salt Na. X. Ka = 4 x 10 -5 mol dm-3. Ka = [H+(aq)] [X¯(aq)] [HX(aq)] re-arrange [H+(aq)] = [HX(aq)] Ka [X¯(aq)]

Calculating the p. H of an acidic buffer solution Calculate the p. H of the solution formed when 500 cm 3 of 0. 1 mol dm-3 of weak acid HX is mixed with 500 cm 3 of a 0. 2 mol dm-3 solution of its salt Na. X. Ka = 4 x 10 -5 mol dm-3. Ka = [H+(aq)] [X¯(aq)] [HX(aq)] re-arrange [H+(aq)] = [HX(aq)] Ka [X¯(aq)] The solutions have been mixed; the volume is now 1 dm 3 therefore [HX] = 0. 05 mol dm-3 [X¯] = 0. 10 mol dm-3 and

Calculating the p. H of an acidic buffer solution Calculate the p. H of the solution formed when 500 cm 3 of 0. 1 mol dm-3 of weak acid HX is mixed with 500 cm 3 of a 0. 2 mol dm-3 solution of its salt Na. X. Ka = 4 x 10 -5 mol dm-3. Ka = [H+(aq)] [X¯(aq)] [HX(aq)] re-arrange [H+(aq)] = [HX(aq)] Ka [X¯(aq)] The solutions have been mixed; the volume is now 1 dm 3 therefore Substituting [HX] = 0. 05 mol dm-3 [X¯] = 0. 10 mol dm-3 [H+(aq)] = 0. 05 x 4 x 10 -5 and = 2 x 10 -5 mol dm-3 = 4. 699 0. 1 p. H = - log 10 [H+(aq)]

SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate.

SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of strong acids and strong bases - SODIUM CHLORIDE Na. Cl dissociates completely in water Water only ionises to a very small extent Na+ Cl¯ H 2 O ——> Na+ + Cl¯ OH¯ + H+ Na+ and OH¯ H+and Cl¯ are ions of a strong base so remain apart are ions of a strong acid so remain apart all the OH¯ and H+ ions remain in solution therefore [H+] = [OH¯] and the solution will be NEUTRAL

SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of strong acids and weak bases - AMMONIUM CHLORIDE NH 4 Cl dissociates completely in water Water only ionises to a very small extent NH 4+ Cl¯ ——> H 2 O NH 4+ + Cl¯ OH¯ + H+ NH 3 + H 2 O Na+ and OH¯ H+and Cl¯ are ions of a strong base so tend to be associated are ions of a strong acid so remain apart all the H+ ions remain in solution therefore [H+] > [OH¯] and the solution will be ACIDIC

SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of weak acids and strong bases - SODIUM ETHANOATE CH 3 COONa dissociates completely in water Water only ionises to a very small extent CH 3 COO ¯ Na + ——> Na+ H 2 O OH¯ + CH 3 COO¯ + H+ CH 3 COOH Na+ and OH¯ H+and CH 3 OO¯ are ions of a strong base so remain apart are ions of a weak acid so tend to be associated all the OH¯ ions remain in solution therefore [OH¯] > [H+] and the solution will be ALKALINE

SALT HYDROLYSIS Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of weak acids and weak bases - AMMONIUM ETHANOATE CH 3 COONH 4 dissociates completely in water CH 3 COO ¯ NH 4 + ——> NH 4+ + CH 3 COO¯ Water only ionises to a very small extent H 2 O OH¯ + H+ NH 3 + H 2 O CH 3 COOH Na+ and OH¯ H+and CH 3 OO¯ are ions of a weak base so tend to be associated are ions of a weak acid so tend to be associated the solution might be alkaline or acidic i. e. APPROXIMATELY NEUTRAL

REVISION CHECK What should you be able to do? Recall the definition of a buffer solution Recall the difference between an acidic and an alkaline buffer solution Recall the uses of buffer solutions Understand the action of buffer solutions Calculate the p. H of an acidic buffer solution Recall and understand the reactions due to salt hydrolysis CAN YOU DO ALL OF THESE? YES NO

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